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3 years ago

Which best describes the electromagnetic force?

Physics

2 answers:

Solnce55 [7]3 years ago

7 0

B.strong but weaker than the weak force

Lunna [17]3 years ago

7 0

C. weak, but stronger than the gravitational force

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A 45.2-kg person is on a barrel ride at an amusement park. She stands on a platform with her back to the barrel wall. The 3.74-m

elena-14-01-66 [18.8K]

Answer:

1,230N

Explanation:

1. Name of the variables:

$f:frequency\\\\ \omega:angular\text{ }speed\\\\ a_c:centripetal\text{ }acceleration\\\\ F_c:centripetal\text{ }force\\ \\ m:mass\\ \\ d:diameter\\ \\ r:radius\\ \\ g:gravitational\text{ }acceleration$

2. Formulae:

$f=\dfrac{number\text{ }of\text{ }revolutions}{time}$

$\omega=2\pi f$

$a_c=\omega^2 r$

$F_c=m\times a_c$

3. Solution (calculations)

$f=\dfrac{1}{1.65s}=0.\overline{60}s^{-1}$

$\omega=2\pi\times0.\overline{60}\approx 3.808rad/s$

$a_c=(3.808rad/s)^2\times (3.74/2m)=27.12m/s^2$

$F_c=45.2kg\times27.12m/s^2=1,225.67N\approx 1,230N$

3 0

3 years ago

What is the effect on the force of gravity between two objects if the mass of one object remains unchanged while the distance to

Vadim26 [7]

Answer:

The correct answer to the question is

B. It always decreases

Explanation:

To solve the question, we note that the foce of gravity is given by

$F_G=\frac{Gm_1m_2}{r^2}$ where

G= Gravitational constant

m₁ = mass of first object

m₂ = mass of second object

r = the distance between both objects

If the mass of one object remains unchanged while the distance to the second object and the second object’s mass are both doubled, we have

$F_{G2} =\frac{Gm_1(2m_2)}{(2r)^2}$ = $\frac{2}{4} \frac{Gm_1m_2}{r^2}$

Therefore the gravitational force is halved. That is it will always decrease

4 0

3 years ago

Let x define the position of an object such that x =

timama [110]

See the attached picture for answers

7 0

2 years ago

For a ∆x of 0.1mm, what is ∆px, the uncertainty in the transverse momentum of a photon passing through a slit (where uncertainty

Colt1911 [192]

Answer:

$0.53\times 10^{-30}kgms^{-1}$

Explanation:

Uncertainty principle say that the position and momentum can not be measured simultaneously except one relation which is described below,

$\Delta x\Delta p=\frac{h}{4\pi }$

Given that the uncertainty in x is 0.1 mm.

Therefore,

$\Delta p=\frac{6.626\times 10^{-34} }{4\times 3.14\times 1\times 10^{-4}m }\\\Delta p=0.53\times 10^{-30}kgms^{-1}$

Therefore, uncertainty in the transverse momentum of photon is $0.53\times 10^{-30}kgms^{-1}$

6 0

3 years ago

It is a general exercise that we do before practicing sports activities worm-up or cool down

Montano1993 [528]

It is called a warm-up.

3 0

3 years ago

Read 2 more answers